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Black_prince [1.1K]
4 years ago
9

What affects the apparent brightness of a star

Physics
1 answer:
Ksivusya [100]4 years ago
4 0
I am not sure, but i believe it is the distance from the viewing.
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Help please!!! Physics circular motion
svp [43]

Answer:

2. 3.1415 m/s

3. 0.63m

4. 0.006 m/s^2

Explanation:

2. v=(2*pi*r))/T, put in values and solve.

3. Circumference=2*pi*radius, radius is 0.1m, plug in and solve.

4. a=(v^2)/r, (don't worry about converting to meters, since units are the same in kilometers, they will cancel out) plug in values and solve.

That's it!

6 0
3 years ago
Wire sizes are often reported using the AWG (American Wire Gauge) system in which smaller diameter wires are said to be of highe
NISA [10]

Answer:

Explanation:

24 - gauge wire , diameter = .51 mm .

Resistivity of copper ρ = 1.72 x 10⁻⁸ ohm-m

R = ρ l / s

1.72x 10⁻⁸ / [3.14 x( .51/2)² x 10⁻⁶ ]

= 8.42 x 10⁻² ohm

= .084 ohm

B )  Current required through this wire

= 12 / .084 A

= 142.85 A

C )

Let required length be l

resistance = .084 l

2 = 12 / .084 l

l = 12 / (2 x .084)

= 71.42 m

6 0
3 years ago
Read 2 more answers
A spy in a speed boat is being chased down a river by government officials in a faster craft. Just as the officials’ boat pulls
babunello [35]

Answer:

They are 7.4m apart.

Explanation:

Here we have a parabolic motion problem. we need the time taken to land so:

Y=Yo+Vo*t+\frac{1}{2}*a*t^2

considerating only the movement on Y axis:

0=4.6-(9.81)*t^2\\t=0.68s

Because we have a contant velocity motion on X axis:

xs=vs*t\\xs=17m/s*(0.68s)\\xs=11.6m

and

xg=vg*t\\xg=28m/s*(0.68s)\\xg=19m

the distance between them is given by:

d=|xg-xs|\\d=7.4m

7 0
3 years ago
What is true of the moon's orbital and rotational periods?
Eddi Din [679]
The moon's orbital and rotational periods are identical or the same, I<span>ts rate of spin is done in unison with its rate of revolution (the time that is needed to complete one orbit). Thus, the moon rotates exactly once every time it circles the Earth.</span>
4 0
4 years ago
Suppose you are measuring double‑slit interference patterns using an optics kit that contains the following options that you can
svetlana [45]

Answer:

3.6 m

Explanation:

\lambda_R = 650 \ nm\\\\\lambda_R = 650*10^{-9} m\\\\L \ should \ be \ minimum \\\\i.e \  0.25 \ mm\\\\= 0.25 *10^{-3} m

\lambda_R = 700 \ nm\\\\\lambda_R = 700*10^{-9} m\\\\

Also

\beta = 1 \ mm \ fringe \  width

D_{min} = \frac{\beta d}{\lambda}\\\\D_{min} = \frac{10^{-3}*0.25*10^{-3}}{700*10^{-9}}\\\\D_{min} = 3.57 \\D_{min} =  3.6 m

Therefore, the minimum distance L  you can place a screen from the double slit that will give you an interference pattern on the screen that you can accurately measure using an ordinary 30 cm (12 in) ruler. = 3.6 m

4 0
4 years ago
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